检测序列的置换 [英] Detect a permutation of sequence

问题描述

我有一个数字是这样的（阵列）名单

I have a list of numbers like this (array)

1 2 3 4

所以我的目标是给另一个数组，如果检查这个数组，如果原来的例子的置换，数组（3 4 1 2）和（1 2 4 3）是原来的，但（1 2 1 1）或（1 5 4排列3）不是。

So my goal is the check given another array if this array if a permutation of the original example, the array (3 4 1 2) and (1 2 4 3) are permutations of the original but (1 2 1 1) or (1 5 4 3) not.

推荐答案

两个可能的解决方案是：

Two possible solutions are:

（1） O（N）空间和放大器;数据的平均时间的解决方案是创建一个直方图，基于哈希表， - 和检查直方图identicals。我们的想法是 - 算多少的每个元素出现在每一个列表中，然后检查每一个元素出现的完全的每个数组中相同的时间

(1) O(n) space & average time solution will be to create a histogram, based on a hash table, of the data - and check if the histograms are identicals. The idea is - count how many each element appears in each list, and then check to see each element appears exactly the same times in each array.

伪code：

map1 = new map //first histogram
map2 = new map //second histogram
for each element in arr1: //create first histogram
   if (element in map1):
         map1.put(element,map1.get(element)+1)
   else:
         map1.put(element,1)
for each element in arr2: //create second histogram
   if (element in map2):
         map2.put(element,map2.get(element)+1)
   else:
         map2.put(element,1)
for each key in map 1: //check all elements in arr1 appear in arr2
   if map1.get(key) != map2.get(key):
        return false
//make sure the sizes also match, it means that each element in arr2 appears in arr1.
return arr1.length == arr2.length 

（2） O（nlogn）时间解决方法是将两个数组排序，然后遍历，并检查它们是相同的。

(2) O(nlogn) time solution will be to sort both arrays, and then iterate and check if they are identical.

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